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Den 19.03.2024 10:17, skrev Richard Hachel:Vous vous trompez.Le 18/03/2024 à 22:12, "Paul B. Andersen" a écrit :No error!A rocket is stationary on Earth, When its clock show τ = 0 andAbsolutely.
the Earth clock show t = 0 the rocket engine starts an give
the rocket a constant proper acceleration a = 10 m/s².
a = 10 m/s² = 1.05265 ly/y/y c = 1 ly/y d = 12 ly
>> According to SR:
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>
The rocket will pass Tau Ceti at the terrestrial time:
t = √((d/c)²+2⋅d/a) = 12.9156 y
The proper time of the rocket when it passes Tau Ceti is:Here is the error.
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
What SR predicts is not a matter of opinion,
it is a matter of fact.
So _ACCORDING TO SR_:
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The proper time of the rocket when it passes Tau Ceti is:
τ = (c/a)⋅arsinh(a⋅t) = 3.13894 y
The speed of the rocket in the terrestrial frame
when it passes Tau Ceti is:
v = a⋅t/√(1 + (a⋅t/c)²) = 0.9973 ly/y
Facts. Indisputable!
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